3
$\begingroup$

The stock price is assumed to evolve as $S_{t}=S_{0}+\mu t+\sigma B_{t}$, where $S_{0}>0, \mu>0$ and the process $B_{t}$ is Brownian motion.

The saving account is assumed to be $\beta_{t}=e^{r t}$, with interest rate $r$

A call option with strike $K$ and expiration $T$ pays $C_{T}=\left(S_{T}-K\right)^{+}$ at time $T$.

Assume r = 0. Give the EMM.

My attempt

I am a bit lost when it comes to EMM but this is what I have so far:

  • Girsanov's theorem:

$B_{t}$ is a B.M under measure P and C is a constant. Then there exists an EMM q such that:

$\hat{B}_{t}=B_{t}+C_{t} \sim Q$ brownian motion.

$d S_{t}=\mu d t+\sigma d B_{t}$

$r=0 \quad c=\frac{\mu-r}{\sigma}=c=\frac{\mu}{\sigma}$

I am not sure how to continue and give the EMM explicitly.

Edit: After some researching, I have found that the EMM does exist for Bachelier model and is unique by Girsanov's theorem. But I am still a bit lost on how to find it.

$\endgroup$

1 Answer 1

1
$\begingroup$

It sounds to me like you understand everything apart from how Girsanov's theorem defines the EMM.

Girsanov's theorem tells us that if $B_t$ is standard Brownian motion under $P$, then for any adapted process $\gamma_t$ (satisfying certain conditions) the process $\hat{B}_t$ defined by:

\begin{equation} d\hat{B}_t = \gamma_t dt +dB_t \end{equation} is Brownian motion under another equivalent measure and this equivalent measure (let's call it $Q$) can be defined by its Radon-Nikodym derivative:

\begin{equation} \frac{dQ_T}{dP} = \exp\bigg\{ -\int_0^T \gamma^\top(s) dB_s - \frac{1}{2} \int_0^T \gamma^2(s) ds \bigg\}. \end{equation}

Now, as you have already noticed, what I am calling $\gamma(t)$ should in our case be equal to $\frac{\mu - r}{\sigma} = \frac{\mu}{\sigma}$. Then $d\hat{B}_t = \frac{\mu}{\sigma} dt +dB_t$ and we have: \begin{equation} dS_t = \sigma d\hat{B}_t \end{equation} as desired. So, our EMM, $Q = Q_T \sim P$, is defined by the Radon-Nikodym derivative: \begin{align} \frac{dQ_T}{dP} &= \exp\bigg\{ -\int_0^T \frac{\mu}{\sigma} dB_s - \frac{1}{2} \int_0^T \frac{\mu^2}{\sigma^2} ds \bigg\} \\ &= \exp \Big\{ -\frac{\mu}{\sigma}B_T - \frac{1}{2} \frac{\mu^2}{\sigma^2}T \Big\} \end{align}

$\endgroup$
2
  • $\begingroup$ Thank you! I am a bit lost when u introduced the Radon-Nikodym derivative. Could you explain how did you get $\frac{d Q_{T}}{d P}=\exp \left\{-\int_{0}^{T} \gamma^{\top}(s) d B_{s}-\frac{1}{2} \int_{0}^{T} \gamma^{2}(s) d s\right\}$ $\endgroup$ May 13, 2021 at 13:58
  • $\begingroup$ @codelearner, see chapter 3 of galton.uchicago.edu/~lalley/Courses/390/Lecture10.pdf . $\endgroup$
    – R. Rayl
    May 13, 2021 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.